What is (\partial)/(\partial y)(2sin(y^2)) ?

The answer to (\partial)/(\partial y)(2sin(y^2)) is 4ycos(y^2)

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(\partial)/(\partial y)(2sin(y^2))

\frac{\partial}{\partial y} (2 sin (y^{2}))

4 y cos (y^{2})

Solution steps

\frac{\partial}{\partial y} (2 sin (y^{2}))

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 2 \frac{\partial}{\partial y} (sin (y^{2}))

Apply the chain rule:

cos (y^{2}) \frac{\partial}{\partial y} (y^{2})

= cos (y^{2}) \frac{\partial}{\partial y} (y^{2})

\frac{\partial}{\partial y} (y^{2}) = 2 y

= 2 cos (y^{2}) \cdot 2 y

Simplify

= 4 y cos (y^{2})

x \to 5 lim (\frac{sin ( x )}{x})

\int_{0}^{3} (x + 2) x + 1 d x

\int \frac{x ^{14}}{x ^{5} - 1} d x

\frac{\partial}{\partial z} (e^{x yz})

\int (e^{3 x} - 4)^{2} d x

What is (\partial)/(\partial y)(2sin(y^2)) ?
The answer to (\partial)/(\partial y)(2sin(y^2)) is 4ycos(y^2)